This is the production of a magnetic field by current flowing in a conductor.

The magnetic effect of current was discovered by Ousted in 1820. The verified magnetic effect of current by the following simple experiment.​​

Figure below shows a conducting wire AB Above a magnetic needle parallel to it.


So long as there is no current in the wire, the magnetic needle remains parallel to the wire i.e. there is no deflection in the magnetic needle.​​

As soon as the current flows through the wire AB, the needle is deflected.


Magnetic needle

When the  current in wire  AB  is  Reversed the  needle is  deflected  in the  opposite  direction​​

This  Deflection is  a convincing proof of  the  existence  of  a magnetic field  around  a  current  carrying conductor.

On increasing the current in the wire AB the deflection of the needle is increased and vice versa.

This  shows  that  magnetic field  strength  increases  with  the  increase in  current and  vice versa  

It is clear from Worsted’s experiment that current carrying conductor produces a magnetic field around it.​​ 

The  larger the  value  of  current in the  conductor the  stronger is the  magnetic  field and  vice  versa.

Magnetic field

Is the  region around a magnet where magnet effect can be experienced.​​


Is the space around a current carrying conducting (magnet) where magnetic effects can be experienced.​​

The  Direction of  a field at a point is  taken to be  the  direction in  which  a  north magnetic pole would  move more  under  the  influence  of  field  if it  were placed at  that point.

The magnetic field is represented by magnetic lines of force which form closed loops.​​

The magnetic field disappears as soon as the current is switched off or charges stop morning.

Magnetic  flux is  a  measure  of  the number  of  magnetic field  lines passing  through the  region.

The  unit of  magnetic flux is  the  Weber (Wb)
The  flux through an  area  A  on figure  below  the   normal  to  which  lies  at  angle ð
œƒ ​​ to ​​ a ​​ field ​​ of ​​ flux ​​ density B​​ 
 ​​  ​​  ​​  E:\..\..\..\thlb\cr\tz\__i__images__i__\Capture_121.PNG

​​           ​​ 
Is  a  quantity  which  measures the  strength  of  the  magnetic field​​

It  is  sometimes  called magnetic  ​​

It is  a vector  quantity​​

The  SI unit  of  Magnetic  flux density is  Tesla (T) or  Wb/m2

Magnetic flux density is simply called magnetic field B​​

B = θ/A


Consider a positive charge +Q moving in a uniform magnetic field E:\..\..\..\thlb\cr\tz\__i__images__i__\711.png

​​ with a velocity​​ E:\..\..\..\thlb\cr\tz\__i__images__i__\812.png

Let the  Angle between 
​​ and E:\..\..\..\thlb\cr\tz\__i__images__i__\711.png
​​ be θ  as  shown​​ 


It  has  been found experimentally  the  magnetic  field  exerts a  force  F on  the  charge.

The  magnitude  F of  this  force  depends  on the  following factors​​ 
(i) F  α θ

(ii)F  α B


Combining the factors we get​​ 

Where K is a constant of proportionality​​

The unit of B is so defined that K = 1​​ 

Equation (a) can be written in a vector form as:-
F = the force of the particle (N)​​ 
B = the magnitude of the magnetic flue density of the field T

Q = the charge on the particle

V= the magnitude of the velocity of the particle​​ 

Definition of ​​ 
F = BQVsinâ
  ​​  If V = 1, Q = 1,​​ θ= 90 then

F = Sin90

F=B     ​​ 

Magnetic field (​​ 
) at a point in space is equal to the force experienced by a unit charge moving with a unit velocity perpendicular to the direction of magnetic field at that point  

Right Hand Grip Rul

Grip the wire  using  the  right hand with the  thumb  pointing in the  direction of the  current  the  other fingers unit point in the  direction of  the  field.

–          For an electron (negatively charged) entering the magnetic field as shown below


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image001.gif

The Direction of positive charge will be exactly opposite. Applying Right hand Grip Rule it is clear that  Direction  of force on the  electron will be  vertically  upward

For a  positively charged particle, it will be  vertically downward

Direction of magnetic field means from N –pole to S-pole.


Consider an electric charge Q moving with a velocity V through a magnetic field B. then the magnetic force F on the charge is given by

                               F = BQVE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image002.gif

(i)        When ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image003.gif

 = 0o​​ or  1800  

F = BQVE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image002.gif

F=   BQVSin00​​ or   F = BQVE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image004.gif

 F= 0                       ​​ 

Hence  a charged particle moving parallel(or Anti parallel) to the  direction of magnetic field experiences no force

(ii)       When ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image003.gif


 F = BQVE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image002.gif

F = BQVE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image005.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image005.gif

 F = BQV

Hence a force experienced by charged particle is maximum when it is moving perpendicular to the direction of magnetic field.

(iii)    When V=O, the charge particle is at rest.

F = BQVE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image006.gif

F=BQ(0)E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image006.gif


If a charged particle is at rest in a magnetic field it experiences no force.

(iv)   When Q = O​​ 

 F = BQVE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image002.gif

F = 0​​ 

Hence electrically neutral particle (eg neutron) moving in a magnetic field experiences no force. 

The magnetic force F acts perpendicular to velocity V (as well as B)

This  means  that  a uniform magnetic field  can  neither  speed  up  nor  slow down a  moving charged particle;  it  can  charge only the  Direction  of V and  not  magnitude of  V​​ 

Since  the  magnitude  of  V does  not  charge the  magnetic force  does not change  the  kinetic energy of  the  charged particle.

UNITS AND DIMENSIONS OF​​ E:\..\..\..\thlb\cr\tz\__i__images__i__\711.png

 The SI unit of magnetic field B is​​ Tesla


            F = BQVE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image002.gif

            ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image007.gif

If   Q = 1C, V =1m/s, Q= 900   F= 1N​​ 

B  = 1T​​ 

 Hence   the  strength of  magnetic field  at a  point is  1T if  a charge of  1C when  moving  with a  velocity of  1m/s  at  right angles to  the  magnetic field, experiences a  force  of  1N at  that  points.

Magnetic field of earth at surface is about 10​​ – 4T. On the other hand, strong electromagnets can produce magnetic fields of the order of 2T.

Dimensions of​​ E:\..\..\..\thlb\cr\tz\__i__images__i__\711.png

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image007.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image008.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image009.gif


Worked Examples

1.       A proton is moving northwards with a velocity of​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image010.gif

m/s in a magnetic field of 0.1Tdirected eastwards. Find the force on the proton. Charge on  proton = 1.6 x 10​​ -19C.


 F = BQVE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image002.gif

B= 0.1T

 V=​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image011.gif

F=0.1 X 1.6 X 10
-19​​ X 5 X 106X Sin 90​​ 

Q = 1.6 X 10-19C

 E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image012.gif

= 900


2.  An electron experiences the greatest force as it travel   at 3.9 x105​​ m/s in a magnetic field when it is moving westward.  The force   is upward and is of magnitude​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image014.gif

N what is the magnitude and direction of the magnetic field.


The conditions of the problem suggest that the electron is moving at right angles​​ 

                 ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image015.gif

 To the direction of the magnetic field

               F = BQVE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image002.gif

, F = 8.7 x 10​​ -13N

                Q= 1.6 X10​​ -19C



              B = 13.14T

By right hand rule per cross product, the direction of the magnetic field is towards northward.

3.    An  α  – particle of mass 6.65 x 10-27​​ kg is  travelling at right angles to a magnetic field with a speed of 6×105m/s. The strength of   the magnetic field is 0.2T.calculate the force on the​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image018.gif

 – particle and its acceleration.

  Force on   α – particle   F = BQVE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image002.gif

  M = 6.65 X10-27Kg​​ 

  V = 6 x 105m/s

   B = 0.2T

  ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image012.gif

 = 900​​ 

    F = BQVE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image006.gif

    = (0.2 x 2x 1.6×10-19) x​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image019.gif

x Sin90Ëš


      Acceleration of α – particle​​ 

     F= mÉ‘​​ 

   É‘=​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image021.gif

=   E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image022.gif

4.       A  copper  wire  has  1.0 x 1029​​ free  electrons per  cubic meter, a  cross sectional  area  of  2mm2​​ and  carries  a  current of  5A  . The wire is placed at right angle to a uniform magnetic field of strength 0.15T. Calculate the force the acting on each electron.


        ​​ I = neAE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image024.gif

Drift velocity =​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image025.gif

n= 1×1029m-3   e = 1.6×10-19c   A= 2mm2​​ = 2×10-6m2

         I = 5A​​ E:\..\..\..\thlb\cr\tz\__i__images__i__\193.png

Force on each electron F= BQE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image024.gif

 SinE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image012.gif

Q= 1.6 x 10-19c

B= 0.15T                                             ​​ 




The  Biot – Savart  law  states  that the  magnitude  of  magnetic  flux  density  dB  at a point  P  which is  at a distance  r  from  a very short  length  dl of  a conductor  carrying  a current I  is  given by.​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image033.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image034.gif

where​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image012.gif

 is  the  Angle between the  short length dl and  the  line  joining  it to point  P

          K  is a constant of  proportionality  its  value  depends on the  medium in which the  conductor is  situated and  the  system of units  adopted.

           For  free space  vacuum  or air​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image039.gif



This equation is known as Biot –Savart Law and gives the magnitude of the magnetic field at a point due to small current element

Current element

 Is  the  product of  current (I)  and  length of  very small  segment (E:\..\..\..\thlb\cr\tz\__i__images__i__\2212.png

) of  the  current carrying  conductor.

Current element =E:\..\..\..\thlb\cr\tz\__i__images__i__\294.png

Current element produces magnetic field just as a stationary charge produces an electric field the current element is a vector.

Its  Direction is  Tangent  to the  element and  acts in the direction of  current flow  in the  conductor​​ 

Biot -Savart law holds strictly per steady currents

Direction of     BE:\..\..\..\thlb\cr\tz\__i__images__i__\332.png

The  direction  of  E:\..\..\..\thlb\cr\tz\__i__images__i__\257.png

  is  perpendicular  to  the  plane  containing   E:\..\..\..\thlb\cr\tz\__i__images__i__\2213.png
   and   E:\..\..\..\thlb\cr\tz\__i__images__i__\267.png
     by  right hand rule  for the  cross  product the  field  is  directed inward. ​​ 

Special cases​​ 

           ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image047.gif

(i)   When​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image012.gif

 = 00​​ or 1800

i.e Point P lies on the axis of the conductor

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image049.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image050.gif

Hence there is no magnetic field at any point on the thin current carrying conductor minimum value.

(ii)    When​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image012.gif

 = 900

When point P lies at a perpendicular position​​ w .r. t​​ current element


Hence magnetic field due to a current element is maximum in a plane passing through the element and perpendicular to its axis.

Important point about Biot – Savant law

(i)   Biot – Savant law is valid per symmetrical current distributions.

(ii)    Biot – Savant  law  cannot  be  proved  experimentally because  it is  not  possible to have  a current  carrying  conductor  of  length dl

(iii)     Like  coulomb’s  law  in  electrostatics, Biot- Savant law  also obeys  inverse square  law

(iv)     The  Direction of  dB   is  perpendicular to  the  plane  containing  E:\..\..\..\thlb\cr\tz\__i__images__i__\295.png

  and ​​ E:\..\..\..\thlb\cr\tz\__i__images__i__\268.png

(v)      This  law  is  also  called  Laplace’s  law and  inverse square law’


 According  to  coulomb’s  law  in  electrostatics, the  eclectic field due to  a  charge  element dQ  at a distance  r is  given by​​ 

       ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image056.gif

According to Biot – Savart law the magnetic field due to a current element  E:\..\..\..\thlb\cr\tz\__i__images__i__\296.png

  at a distance r is given by  ​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image058.gif

From the above two equations we note the following points of Similarities and Dissimilarities.


(i)   Both laws obey inverse square  law​​ 

(ii)  Both the  fields(magnetic field and  Electro static field) obey  superposition principles

(iii)Both the fields are long range fields.


(i)  The Electric field is produced by a scalar source i.e.  Electric chargeE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image060.gif

. However the magnetic field is product by a vector source i.e.  current  E:\..\..\..\thlb\cr\tz\__i__images__i__\296.png

(ii)  The Direction the Electric field is along the displacement vector i.e.  The line joining the source and field point. However  the  direction of  magnetic  field  is  perpendicular  to the  plane  containing current  element ​​ E:\..\..\..\thlb\cr\tz\__i__images__i__\296.png

and  displacement vector  E:\..\..\..\thlb\cr\tz\__i__images__i__\2610.png

(iii)    In Biot –Savant law the magnitude of magnetic field dB α SinE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image012.gif

 Where​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image012.gif
 is the  Angle  between current element E:\..\..\..\thlb\cr\tz\__i__images__i__\296.png
  and  displacement vector   E:\..\..\..\thlb\cr\tz\__i__images__i__\2610.png
  However there is  no  angle  dependence in  coulomb’s law for electrostatics


 Consider a circular coil of radius r and carrying current I in the Direction shown in figure​​ 


                                                                  ​​ E:\..\..\..\thlb\cr\tz\__i__images__i__\302.png

Suppose  the   loop  lies  in the  plane  of paper it is  desired  to find  the  magnetic field  at  the  centre O of  the  coil​​ 

Suppose  the  entire  circular coil is divided into a  large  number  of  current  elements each of  length​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image038.gif

According  to  Biot – Savant  law, the  magnetic field E:\..\..\..\thlb\cr\tz\__i__images__i__\258.png

​​ at the centre O of the  coil  due  to current  element​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image042.gif
 is  given  by​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image058.gif

……………E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image065.gif

The  direction  of  dB  is  perpendicular to the  plane  of the  coil and is  Directed  inwards

Since  each  current  element  contributes  to the magnetic field  in the  same  direction, the  total magnetic field B  at the  centre O can be  found  by integrating equation…………(i)​​ 


              L- Total length of the coil = 2E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image073.gif



 If the coil has N turns each carrying current in the same direction then contribution of all turn are added up.

 B=E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image076.gif



 The flux density dB at P due to the start length dl given by equation as
E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image058.gif

From the figure (A)​​ 

      ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image098.gif

   ,​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image099.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image100.gif
      ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image101.gif

      r =  ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image102.gif


       ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image103.gif

 = a cot​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image104.gif

      ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image105.gif

 = -aE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image106.gif

Substituting for​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image037.gif

 and​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image038.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image107.gif

       ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image108.gif

  The total flux density B at P is the sum of the flux densities of all the short lengths and can be found by letting dE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image012.gif

→O and integrating over the whole length of the conductor.

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image109.gif

      ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image110.gif

 The  limits  of  the  integration  are​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image111.gif

 and 0 because  these are  values of 𝜃​​ at the ​​ ends of the ​​ conductor
    ​​   ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image112.gif

       ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image113.gif

       ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image114.gif

      ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image115.gif



Circular coil with its plane perpendicular to that of the paper

The  flux  density dB at p due  to the  short length dl of the coil  at  X, where  X is  in  the  plane  of the  paper, is  given by  equation as​​ 

   ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image118.gif

          By symmetry, when all the short lengths​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image038.gif

 are taken into account the components of magnitude​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image119.gif
 sum to zero.

          Each  short length  produces a component of magnitude​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image036.gif

Sin α parallel to the  axis and  all those components are  in the  direction shown

          The  total  flux density  is  therefore  in  the  direction of​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image036.gif

Sin α  and  its magnitude B is  given by​​ 

             ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image120.gif

          ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image121.gif

 The radius vector XP of each small length is perpendicular to it, so that​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image012.gif

=900​​ and there pore SinE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image012.gif
 = 1

        ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image122.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image123.gif

= 2E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image124.gif
(the circumference of the coil)

 E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image125.gif

 E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image126.gif

, But​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image127.gif
 =​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image128.gif

 E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image129.gif

 E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image130.gif

For a coil of N Turns

 E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image131.gif

 When S= r​​ 
E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image132.gif

Also from the figure



States that the line integral of magnetic field E:\..\..\..\thlb\cr\tz\__i__images__i__\393.png

   around any closed path in vacuum/air is equal to​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image134.gif
times the total current (I) enclosed by that path​​ 
     ​​ E:\..\..\..\thlb\cr\tz\__i__images__i__\Capture_13.PNG

I = current enclosed by that path.                                 

Ampere’s  law is  an  alternative  to  Biot –  Savart law  but  it is  useful for  calculating  magnetic field  only in situations with considerable symmetry.

This law is true for steady currents only.

In order  to  use  law  it is  necessary  to  choose  a  path  for which it  is possible  to determine the  value of  the  line  integral

It  is  because  there  are  many  situations where there  is  no such path  that  the law is of  limited use.

 Hence the application of ampere law​​

(i) Magnetic field due to constraining conductor carrying current
(ii)Magnetic field due to solenoid carrying current

(iii)Magnetic field due toroid


 Consider a long straight conductor carrying current I in the direction as shown in the figure below

It is desired to find the magnetic field at a point p at a perpendicular distance r for the conductors


Applying Ampere’s circuital law to this closed path​​ 

Is a long coil of wire consisting of closely packed loops


 Is a cylindrical coil having many numbers of turns ​​ 

By  long  solenoid we  mean that  the  length of  the Solenoid is very large as  compared to  its  Diameter.

Figure  below  shows the  magnetic field lines due to an  air cored solenoid carrying current

Inside the solenoid the magnetic field is uniform and parallel to the solenoid axis.​​ 

Outside  solenoid  the  magnetic field is  very  small as  compared  to the  field inside  and  may be  assumed  zero.

It  is  because the  same  no  of  field  line  that  are  concentrated  inside the  solenoid spread out  into very  faster  space  outside​​ 

 Magnetic flux density due to an Axis of an in finely long Solenoid

Consider  the  magnetic  flux  density​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image036.gif

 at P  due  to  a section of the  solenoid of  length ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image149.gif

 n = number of turns per unit length.

N= number  of  turns  the  section can be  treated  as a plane  circular coil of  N turns  in  which  case  dB is  given by​​ 


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image151.gif

Since​​ dx​​ is small, the section can be treated as a plane circular coil or N turns in which case dB is given by           ​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image152.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image153.gif

From the figure​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image154.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image155.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image156.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image157.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image158.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image159.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image160.gif

Substituting for​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image161.gif

 and dx gives,

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image162.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image163.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image164.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image165.gif

The  flux  densities   at  P due to  every  section  of the  Solenoid  are  all  in the  same  direction  and  therefore  the  total  flux  density  B can  be  found by  letting  dB→o and  integrate over  the  whole  length of the  solenoid.​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image167.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image168.gif

The limits of integration are​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image111.gif

 and 0 because these values of β at the end of the solenoid.​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image169.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image170.gif

 E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image171.gif

 E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image172.gif


If the Solenoid is Iron-cored of relatively permeability​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image175.gif

 magnitude of magnetic field inside the Solenoid is​​ 


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image176.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image177.gif


At points near the ends of an air cored Solenoid, the magnitude of magnetic field is​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image179.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image180.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image181.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image182.gif


The magnetic field outside a solenoid is zero

Also in a current carrying long solenoid the magnetic field produced does not depend upon radius of the Solenoid.


Toroid is a solenoid that bent into the form of the closed ring.

The magnitude field B has a constant magnitude every where inside the toroid while it is zero in the open space interior and exterior to the toroid.

If any closed path is inside the inner edge of the toroid then ther is no current enclosed. Therefore, by Ampere’s circular law B=0.

Magnetic field 
​​ due to toroid
Consider the diagram below

                    ​​ E:\..\..\..\thlb\cr\tz\__i__images__i__\98.png

Let r = mean radius of toroid

I = Current through toroid

n = number of turns permit length

B = magnitude of magnetic field inside the toroid







We know that a moving charge in a magnetic field experiences a force

Now electric current in a conductor is due to the drifting of the force electrons in a definite direction in the conductor

When such a current carrying conductor is placed in a uniform magnetic field, each free electron experiences a force.

 Since the free electrons are constrained in the conductor, the conductor itself experiences a force.

          Hence a current carrying conductor placed in magnetic field experiences a force F.

 Consider a conductor of the length L and area of cross- section a placed at an angle 𝜃​​ to the direction of uniform of magnetic field B.

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image012.gif

– is the angle between the plane of the conductor. The magnetic force experienced by the moving charge in a conductor is F = BQV SinE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image012.gif

            For steady current       I =​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image185.gif

                                                Q =I t

                                                F=​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image186.gif

The velocity for direct current is constant

                                                V=​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image187.gif

                                                F = B I tE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image188.gif

                                               ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image189.gif

F= Force on the conductor (N)

B= Magnitude of the magnetic flux density of the field (T)

I = Current in the Conductor (A)​​ 

L= length of the conductor (M)

  The current in the conductor I​​ 


Special cases​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image191.gif

 E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image192.gif
 E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image193.gif

Thus if current carrying conductor is placed parallel to the direction of the magnetic field of the conductor will experience no force.

ii.)​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image194.gif

                        F = BILE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image195.gif

                     ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image196.gif

Hence current carrying conductor will experience maximum force when it placed at right angles to the direction of the field.

One Tesla​​ 

Is the  magnetic  flux density  of a  field  in  which a  force of  IN  acts on  a 1M length of a  conductor  which is  carrying a  current of  IA and  is  perpendicular to  the  field.​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image197.gif

 B = Tesla​​ 


 The Direction of the force​​ 

Experiment shows that  the  force  is  always perpendicular  to the  plane  which  contains both the  current and  the  external field   at the  site of  the  conductor​​ 

The  direction  of the  force  can be  found by  using  Fleming’s left  hand rule​​ 

 Fleming’s left hand rule​​ 

States that if the first and the second fingers and the thumb of the left hand are placed comfortably at right angles to each other, with the first finger pointing in the direction of the current then thumb points in the direction of the force i.e. Direction in which Motion takes place If the conductor is free to move.

Maxwell’s Corkscrew rule

States that if a right handed corkscrew is turned so that its point travels along the direction, the direction of rotation of corkscrew gives the direction of the magnetic field.


When two parallel current carrying conductors are close together, they exert force on each other.

It is because one current carrying conductor is placed in the magnetic field of the other

If currents are in the same direction the conductor attract each other and  If currents are in the opposite directions conductors  repel each other

Thus like currents attract, unlike currents repel.

Consider two infinitely long straight parallel conductors X and Y carrying currents I1​​ and I2​​ respectively in the same direction.


 Suppose the conductors are separated by a distance​​ rin the plane of the paper.

As each conductor is in the magnetic field produced by the other, therefore each conductor experiences a force

The current carrying conductor Y is placed in the magnetic field produced by conductor X

Therefore force act on the conductor Y.  The magnitude of the magnetic field at any point P on the conductor Y due to current I, in the conductor X is

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image200.gif

By  right  hand  grip rule ; the  direction of  B  is  perpendicular  to the  place  of the  paper  and  is  directed  inwards.

 Now conductor Y carrying current​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image201.gif

 is placed in the magnetic field​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image202.gif
produced by conductor X

         Therefore force per unit length of conductor  Y will experience a force​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image203.gif

 given by​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image204.gif

 E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image205.gif

=​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image206.gif

 According to FLHR, force​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image207.gif

 on conductor Y acts in the place of the paper perpendicular to Y and is directed towards to the conductor X.

Similarly, the Force on conductor X per unit length is  ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image208.gif

= ByI1L



E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image209.gif


Hence when two long parallel conductors carry currents in the same direction they attract each other. The force of attraction per unit length is      ​​ 
E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image211.gif



This shows that the attraction between two parallel straight conductors carrying currents in the same direction in terms of magnetic field lines of conductors

It is clear that in the space between X and Y the two fields are in opposition and hence they tend to cancel each other​​ 

However in the space outside X and Y the two fields assist each other. Hence resultant  field  distribution will be​​ E:\..\..\..\thlb\cr\tz\__i__images__i__\F6_T3_electromagnetism.png

If  two straight  current  carrying  conductors of  unequal length  are  held parallel to each other  then force  on the  long  conductor  is  due to the   magnetic field  of the  short conductor​​ 

I1​​ = Current through short conductor​​ 

l​​ = Length of short conductor​​ 

I2​​ = Current through long conductor​​ 

L = Length of long conductor​​ 

If r is the separation distance between these parallel conductors​​ 

Force on Long conductor = force on short conductor​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image214.gif

Force on each conductor is the same in magnitude but opposite in direction (Newton’s third law)


Force between two current currying conductors per unit length​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image215.gif

If ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image216.gif

 And r =1m then



 Is  that  steady  current  which when it is  flowing  in  each  of  two infinitely  long, straight parallel  conductors  which  have  negligible  areas  of  cross – section  and  are  1m apart  in a vacuum, causes each conducts to  exert  a force  of ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image219.gif

N on each mete  of the  other. ​​ 


1. The  plane  of a  circular  coil  is  horizontal  it  has  20 turns  each of  8cm radius A current  of  1A flows  through it  which  appears  to be  clockwise from a point vertically  above  it. Find the  Magnitude  of the  magnetic  field  at the  centre of the  coil.


The  magnitude  of the  magnetic  field  at the  centre of the  coil  carrying  current  is  given by,

As the currents appears to be clockwise from appoint vertically above the coil the direction of the field will be vertically downward (By R.H.G.R)

 ​2. A wire placed along the South-North direction carries currents of 5A from South to North. Find the magnetic field  due to a 1cm piece of wire at a point 200cm North-East from the place.



By RHGR, The field is vertically vertical downwards​​ 

3. A coil of radius 10cm and having 20 turns carries a current of 12A in a clockwise direction when seen from east. The coil is in North – South plane.  Find the magnetic field at the centre of the coil.                                                 ​​ 


The magnitude of the magnetic field at the centre of the coil​​ 


The electron of hydrogen atom moves along a circular path of radius 0.5 x 10-10​​ with the uniform speed of 4 x 106​​ m/s.  Calculate the magnetic field produced by electron at the centre ( e= 1.6 x 10-9c)​​ 

Number the revolution made by the electron in 1 second is

                                      ​​ E:\..\..\..\thlb\cr\tz\__i__images__i__\2215.png

Current   =​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image230.gif

I =  ​​ 1.27 X 1016​​ X 1.6 X10-19


           I =     2.04 X 10-3A

 Magnetic field produced by the electron at the centre is​​ 


5.  A circular  coil  of  100 turns  has  a radius of  10cm and  carries  a current of  5A Determine  the  magnetic  field​​ 

(i)  At  the  centre  of  the  coil​​ 

(ii)  At a point  on the  axis  of  the  coil  at  a distance  of  5cm from the  centre  of the  coil.​​ 


(i)   Magnetic field  at the  centre  of the  coil is​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image233.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image040.gif

 =   4E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image234.gif
 x 10-7​​ TA​​ -1

N    = 100 turns

 I = 5A

 r = 10×10-2m​​ 

B =​​ 4E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image235.gif

 x 10-7x 100 x S

2 X 0.1​​ 

B=   3.14 X10-3​​ T              E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image236.gif

  The magnetic field of the centre of the coil  =   3.14 X10-3​​ T

(ii)    Magnetic  field  on the  axis  of the  coil  at  a  distance  X from the  centre is ​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image133.gif

 E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image040.gif

 = 4E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image234.gif
 x 10-7​​ TA​​ -1​​ 

 N   = 100 turns

  I    = 5A

  r    = 10 x 10-2

  x    = 0.05m​​ 


6. An electric  current  I   is  flowing  in  a  circular  wire  of  radius  at  what  dose  from the  centre  on the  axis  of  circular wire  will the  magnetic field  be  1/8th​​ of its  value  at  the  centre?


Magnetic field B at the centre of the circular coil is​​ 
E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image239.gif

Suppose at a distance X from the centre on the axis of the circular coil the magnetic field is ​​ 
E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image240.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image241.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image242.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image243.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image244.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image245.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image246.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image247.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image248.gif

        7. In  Bohr’s  model of  hydrogen  atom  the  electron  circulates  around  nucleus on a  path of radius  0.51Å at  a  frequency  of  6.8xE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image249.gif

 is  rev/second  calculate  the  magnetic field induction at the  centre of the  orbit.


 The circulating electron is equivalent to circular current loop carrying current I given by​​ 


    ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image250.gif

    ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image251.gif

    I = 1.6​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image252.gif

     I = 1.1E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image253.gif



Magnetic field at the centre due to this current is​​ 


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image254.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image255.gif


                       E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image256.gif

    = 14T​​ 


      8. A long straight wire carries a current of 50A. An  electron  moving at  107ms is  5cm from  the  wire​​ 

                                 ​​ E:\..\..\..\thlb\cr\tz\__i__images__i__\2510.png

 Find the Magnetic field acting on the electron velocity is directed​​ 

(i)    Towards the wire

(ii) Parallel to the  wire​​ 

(iii)  Perpendicular  to the  directions  defined  by  I and  ii


 Solution ​​ 

The magnetic field produced by current carrying long wire at a distance r

 E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image147.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image259.gif

 E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image260.gif

 The field is directed downward perpendicular   to the plane of the paper​​ 

( i)    The velocity V1​​  is towards the wire. The  angle between  VI  and  B is  900​​ force on  electron​​ 

F= BQVE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image006.gif

F = 2x 10-4​​ x 1.6×10-19x107x Sin 900​​ 
F = 3.2 x 10-16​​ N


(ii)  When  the  electron  is  moving  is  moving  parallel to the  wire ,angle  between  V2 and B is  again  90Ëš Therefore, force  is  again ​​ 


(iii)   When  the  electron is  moving perpendicular to the  directions  defined by  (i) and (ii) the  angle  between  V and B is O

F = O


       9. A solenoid  has  a length  of  1 .23 m and  inner diameter  4cm it  has  five  layers of  windings  of  850 turns each and  carries  a current of  5.57A. what  is  the  magnitude  of the  magnetic  field  at the  centre  of the  solenoid​​ 




The  magnitude  of the  magnetic  field at the  centre  of  a  solenoid  is  given  by
E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image262.gif


    10.   A  to void has a  core ( non –  ferromagnetic) of  inner  radius  20cm and  over  radius  25cm around which 1500 turns  of a wire  are  wound. If  current  in the  wire  is  2A

Calculate the magnetic field​​ 

(i)                  Inside  the  to void ​​ 

(ii)               Outside the  to void​​ 


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image267.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image268.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image269.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image270.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image271.gif


( i)   The  magnitude  of the  magnetic  field  inside  the  toroid is  given  by​​ 
E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image262.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image273.gif
    B = 0.003T

(ii)The magnetic field outside the toroid is Zero. It is all inside the toroid.

    11.   A solenoid 1.5m long and 4cm in diameter possess 10 turnsE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image274.gif

cm. A current of 5A is flowing through it. Calculate the  magnetic  induction

(i)   Inside  and​​ 

(ii)  At  one end  on the  axis  of the  solenoid


n =E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image275.gif

 =​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image276.gif
 =​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image277.gif


(i)   ​​ Inside the solenoid , the magnetic induction is given by

B =​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image278.gif

                B = 4E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image279.gif

B =​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image280.gif


(ii)    At the  end  of the  solenoid  the  magnetic  induction is  given  by​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image281.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image282.gif

     12.    (a)  How will the magnetic field intensity at the centre of a circular loop carrying current change, if the current through the coil is doubled and the radius of the coil is halved?


 (b) A long wire first bent in to a circular coil of one turn and then into a circular​​ 

          coil    of  smaller radius having  n  turns, if the  same  current passes in both the  cases, find  the  ratio of  magnetic  fields produced at the  centers in the  two cases.


(c) A and B are  two concentric coils of centre O and carry currents IA​​ and IB  as shown in figure​​ 

                               ​​ E:\..\..\..\thlb\cr\tz\__i__images__i__\285.png

If the ratio of their radii is 1:2 and ratio of flux densities at O due to A and B is 1:3, find the value of​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image284.gif


(a)      Magnetic field at the centre of circular coil

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image285.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image286.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image287.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image288.gif



E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image289.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image290.gif


         ( b)Suppose r is the radius of one turn coil and the r1​​ is the radius of n-turn coil. Then​​ 

NE:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image294.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image295.gif



First case                                  Second case                                        ​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image296.gif

                                  ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image297.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image298.gif


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image299.gif



C.  Magnetic field at the centre of circular coil

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image301.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image302.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image303.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image304.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image305.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image306.gif


     13.   A helium nucleus makes a full rotation in a circle of radius 0.8m in two seconds. Find the value of magnetic field at the centre of the circle.​​ 



The charge on helium nucleus

Q=​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image307.gif


Q=​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image307.gif

 E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image308.gif
 1.6 X10-19c

Current produced I =​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image309.gif

I =​​ 2 x 1.6 x 10-19


I =1.6 x10-19A

Magnetic field at the centre of the circle orbit of the helium is,

  ​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image310.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image311.gif

B= 1.256 x 10-25T  ​​ 

     14.   A soft Iron ring has a mean diameter of 0.20m and an area of cross section 5×10-4m2​​ it is uniformly wound with 2000turns carrying a current of 2A and the magnetic  flux in the iron is 8x 10-3Wb. What is the relative permeability of iron?​​ 



Length of ring l

l = 2E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image312.gif

l = 2E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image312.gif

 x 0.10m


Number of turns per unit length n​​ 

n =​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image313.gif

=​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image314.gif


If M is the absolute permeability of iron, then magnetic flux density of iron ring is​​ 

B =​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image315.gif

B =​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image316.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image317.gif


 Magnetic flux​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image318.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image319.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image320.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image321.gif

 Magnetic flux​​ E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image322.gif

 = BA​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image323.gif

 Relative permeability of Iron μr


E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image324.gif

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image325.gif

     15.   Two  flat  circular  coils  are  made  of  two  identical  wires  each of  length 20cm one  coil  has  number  of  turns  4  and  the  other  2. If the some   current flows though the wire in which will magnetic field at the centre will be greater?



 For the first coil​​ 

E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image326.gif

For second coil​​ 

 E:\..\..\..\thlb\cr\tz\Right Hand Grip Rule_files\image327.gif